Subdifferential boundary conditions: Testing with $L^2$ or $H^{1/2}$ functions

Question

My question was essentially this: Does it make a difference if I test subdifferential boundary conditions with functions from $L^2(\Gamma)$ or $H^{1/2}(\Gamma)$?

In the following, I will phrase the question more precisely after introducing the necessary notation and providing a motivation.

Let $\Omega \subset \mathbb R^d$ be a domain and $\Gamma$ be its boundary. Assume also that we have a convex (proper, lower semicontinuous) function $\phi \colon \mathbb R^d \to \mathbb R \cup \{\infty\}$.

Now we can define a convex functional $j \colon L^2(\Gamma)^d \to \mathbb R$ via $$ j(v) = \begin{cases} \int_\Gamma \phi(v(x))\,dx & \text{if $\phi(v) \in L^1(\Gamma)$,}\\ \infty & \text{otherwise.} \end{cases} $$ For a function $f \in L^2(\Gamma)$ to lie in the subdifferential of this functional $j$ at a point $u$ is by definition equivalent to the condition $$ \tag A f \in L^2(\Gamma) \colon \quad j(v) \ge j(u) + (f,v-u) \quad \forall v \in L^2(\Gamma) $$ Now one might want to restrict $j$ to the subspace $H^{1/2}(\Gamma)^d \subset L^2(\Gamma)^d$ and consider a subdifferential inclusion for this restriction, yielding the condition $$ \tag B f \in H^{-1/2}(\Gamma) \colon \quad j(v) \ge j(u) + (f,v-u) \quad \forall v \in H^{1/2}(\Gamma) $$ Obviously, even if $u$ lies in $H^{1/2}(\Gamma)$, this condition is a weakening of (A) for two reasons:

1. We allow $f$ to be chosen from a larger space.
2. We test with functions from a smaller space.

If we assume $f \in L^2(\Gamma)$, the only difference between (A) and (B) is the condition (2).

Now my question was: Is this weakening strict?

The writing of P.D. Panagiotopoulos suggests that this is not so, but no proof is given.

Update:

@dmw64 has kindly provided a proof sketch for the equivalence of (A) and (B), Thank you! I'll fill in the details here.

The construction is clear; a function $P_n$ can be defined via $$ P_n(x) = \begin{cases} a_n & \text{if $x \le a_n$}\\ x & \text{if $a_n \le x \le b_n$}\\ b_n & \text{if $b_n \le x$.} \end{cases} $$ By redefining $A_n$ if necessary (through an index shift), we can assume $A_n \subset A$ for all $n \ge 1$.

The following claims are now made (I will write $v_n$ for $P_n v$):

1. As $n \to \infty$, we have $\|v_n - v\|_{L^2} \to 0$;
2. Moveover, $\|\phi \circ v_n - \phi \circ v\|_{L^1} \to 0$.
3. For fixed $n$, we can find a sequence $(w_k)$ with $w_k \in H^{1/2}$ such that $\| w_k - v_n \|_{L^2} \to 0$ and $\|\phi \circ w_k - \phi \circ v_n \| \to 0$ as $k \to \infty$.

Before proceeding to the details, here are some notes:

• We can assume $\phi \ge 0$ without loss of generality: Any proper convex function has at least one affine minorant $m$. If we substract this minorant, the result $\phi - m$ still convex, and also non-negative (The additional operator $v \mapsto m \circ v$ that we now need to treat is clearly $L^2$-$L^2$ continuous).
• We have $\phi([y,z]) \le \max(\phi(y),\phi(z))$ for any $y$ and $z$ by convexity of $\phi$.

Now the details:

1. For almost every $x$, the term $|v_n(x) - v(x)|^2$ is monotonically nonincreasing in $n$ and convergent to 0. By the monotone convergence theorem, we thus have $\|v_n -v\|_{L_2} \to 0$ as $n \to \infty$.
2. By construction, we have $v_n(x) \in \operatorname{co}(A_1 \cup \{v(x)\})$ for any $n \ge 1$. This implies $\phi(v_n(x)) \in \phi(\operatorname{co}(A_1 \cup v(x)) \le \max\{\phi(a_1),\phi(b_1),\phi(v(x))\}$. Consequently, we have an integrable bound on $\phi \circ v_n$. Since we obviously have pointwise convergence of $\phi \circ v_n$ to $\phi \circ v$, Lebesgue's convergence theorem now kicks in.
3. Note that $\phi \circ P_n$ is a continuous bounded function. If we define $j_n = \int \phi \circ P_n \circ (\cdot)$, then $j_n \colon L^2 \to L^1$ is continuous by Krasnosel'skii's theorem. We can now find $z_k \in H^{1/2}$ with $\|z_k - v_n\|_{L^2} \to 0$ as $k \to 0$; their truncations $w_k = P_n z_k$ do exactly what we want: On the one hand, they still lie in $H^{1/2}$ (since $P_n$ is Lipschitz); on the other hand, we clearly have $$ w_k = P_n \circ z_k \to P_n \circ v_n = v_n \quad \text{in $L^2$ (as $k \to \infty$)} $$ (again since $P_n$ is Lipschitz) and also $$j(w_k) = j_n(w_k) \to j_n(v_n) = j(v_n)\text. $$ by continuity of $j_n$.

Answer 1

I think that if $f\in L^2$ and $u\in H^{1/2}$, then (B) indeed implies (A). (Note that the formulation of (B) in this form requires $u\in H^{1/2}$). (I omit the $\Gamma$ for brevity... Sorry, it is too late and I am to lazy ;-)...)

So it suffices to show that for every $v\in L^2$ with $j(v)<\infty$ there exists a sequence $w_n\in H^{1/2}$ such that $w_n\rightarrow v$ in $L^2$ and $j(w_n)\rightarrow j(v)$. Then we can pass to the limit using $w_n$ and conclude (A).

Now, here is another sketch. Let $A:={\rm dom\>}\phi$. Then $A$ is an interval in $\mathbb{R}$. If $A$ only consists of one point, then $u=v$ are constant and, in particular, $v\in H^{1/2}$. Therefore, assume that $A$ is not singelton. Let us choose a sequence $A_n=[a_n,b_n]$ of bounded, closed subintervals of $A$ as follows: If $A$ is "left closed" ($A$ is of the form $A=[a,b)$ of $A=[a,b]$), then take $a_n=a$, choose $a_n=a+\frac 1 n$ otherwise and if $a>\infty$ and $a_n=-n$ if $a=\infty$. Similarly, we proceed with $b_n$. Then $A_n$ is a subinterval of $A$ if $n$ is sufficiently large and $\phi$ is bounded and continuous on $A_n$. Now, $\phi\circ P_n v$ should converges to $\phi\circ v$ in $L^1$ and hence $j(P_n v)$ to $j(v)$. Here, $P_n$ is the projection of $\mathbb{R}$ onto $A_n$. Moreover, $P_n v$ converges to $v$ in $v$ in $L^2$ and point-wise almost everywhere ($v$ only takes values only in $A$ a.e.). But we can find a function $w_n\in H^{1/2}$ (or even more regular) taking values only in $A_n$ a.e. such that $w_n$ is close to $P_n v$ in $L^2$ and $\phi\circ w_n$ close to $\phi\circ P_n v$ in $L^1$ (since $\phi$ is bounded and continuous on $A_n$). Now, we can pass to the limit.